Find the average rate of change for $f(t)=e^t-13$ from $t=0$ to $t=5$. Leave your answer in exact form. Do not use decimals.
The average rate of change is $\square$.
Answer (1)
Calculate f ( 5 ) and f ( 0 ) using the function f ( t ) = e t − 13 .
Find f ( 5 ) = e 5 − 13 and f ( 0 ) = e 0 − 13 = − 12 .
Apply the average rate of change formula: 5 − 0 f ( 5 ) − f ( 0 ) .
Simplify to get the average rate of change: 5 e 5 − 1 .
Explanation
Understanding the Problem We are asked to find the average rate of change of the function f ( t ) = e t − 13 from t = 0 to t = 5 . The average rate of change is a measure of how much the function changes per unit of t over the given interval.
Formula for Average Rate of Change To find the average rate of change, we need to calculate the function's value at the endpoints of the interval, which are t = 0 and t = 5 . Then, we'll use the formula for the average rate of change: Average Rate of Change = 5 − 0 f ( 5 ) − f ( 0 )
Calculating Function Values First, let's calculate f ( 5 ) : f ( 5 ) = e 5 − 13 Next, let's calculate f ( 0 ) : f ( 0 ) = e 0 − 13 = 1 − 13 = − 12
Substituting Values into the Formula Now, we can plug these values into the formula for the average rate of change: Average Rate of Change = 5 − 0 ( e 5 − 13 ) − ( − 12 ) = 5 e 5 − 13 + 12 = 5 e 5 − 1
Final Answer Therefore, the average rate of change of f ( t ) = e t − 13 from t = 0 to t = 5 is 5 e 5 − 1 .
Examples
The average rate of change is a fundamental concept in calculus with numerous real-world applications. For example, consider a population of bacteria growing exponentially. The function f ( t ) = e t − 13 could model the number of bacteria at time t (in hours), with the -13 representing some initial condition or offset. Finding the average rate of change between two time points, say t = 0 and t = 5 , tells us the average growth rate of the bacteria population over that 5-hour period. This information is crucial in fields like microbiology and medicine for understanding and controlling bacterial growth.
